Metals and Their Compounds Practice Questions — JAMB UTME — Page 3

Question 41Question →

Copper(II) oxide ( $\text{CuO}$ ) is a black solid compound formed when copper metal is strongly heated in air. Given the relative atomic masses of copper ( $\text{Cu} = 64$ ) and oxygen ( $\text{O} = 16$ ), what is the percentage by mass of copper in pure copper(II) oxide?

Show answer & explanation

Answer: 80

Answer

The percentage by mass of copper in pure copper(II) oxide (

\text{CuO}

) is

80\%

.

The molar mass of copper(II) oxide (

\text{CuO}

) is

64 + 16 = 80\text{ g/mol}

. The relative mass contributed by copper is

64\text{ g/mol}

. Dividing

64

by

80

and multiplying by

100

gives

80\%

.

Step-by-Step Solution

1

Calculate the molar mass of copper(II) oxide (

\text{CuO}

).

Molar mass of

\text{CuO} = 64 + 16 = 80\text{ g/mol}

.

The molar mass of a binary compound is the sum of the relative atomic masses of its constituent elements.

2

Calculate the mass percentage of copper in the compound.

\text{Percentage of Cu} = \left(\frac{64}{80}\right) \times 100\% = 80\%

.

The mass percentage is found by dividing the mass contributed by copper by the total molar mass of the compound and multiplying by 100.

Key Concept

Percentage composition by mass of an element in a copper compound

Question 42Question →

During the industrial extraction of aluminium by the Hall-Héroult electrolytic process, why must the graphite anode rods be replaced at regular intervals?

The oxygen gas discharged at the anode reacts with the graphite at high operating temperatures to form carbon dioxide

Molten aluminium is deposited directly onto the anode rods, causing severe corrosion and electrical shorting

Molten cryolite reacts chemically with graphite to produce gaseous carbon tetrafluoride

Graphite oxidizes directly by releasing carbon cations into the molten electrolyte solution

Show answer & explanation

Answer: The oxygen gas discharged at the anode reacts with the graphite at high operating temperatures to form carbon dioxide

Answer

The oxygen gas discharged at the anode reacts with the graphite at high operating temperatures to form carbon dioxide

During the Hall-Héroult process, oxide ions (

O^{2-}

) are discharged at the carbon (graphite) anode to produce oxygen gas. Because the cell operates at approximately 950°C, the freshly liberated oxygen gas reacts with the carbon anodes to form carbon dioxide (

CO_2

) and carbon monoxide (

CO

). As a result, the graphite anodes are burned away over time and must be routinely replaced.

Step-by-Step Solution

1

Identify the anode half-reaction during the electrolysis of molten alumina in cryolite

Oxide ions (

O^{2-}

) migrate to the positive graphite anode and undergo oxidation to form oxygen gas:

2O^{2-} \rightarrow O_2(g) + 4e^-

Anodes are the site of oxidation where negative ions discharge electrons.

2

Analyze the chemical interaction between liberated oxygen gas and the carbon anode at high cell operating temperatures

At temperatures around 950°C, the liberated oxygen gas reacts with the hot graphite anode to form carbon dioxide:

C(s) + O_2(g) \rightarrow CO_2(g)

Graphite carbon combusts in the presence of oxygen gas at elevated temperatures, leading to continuous depletion of the anodes.

Key Concept

Anode consumption due to oxygen reaction in the Hall-Héroult process

Estimated Time:1m 0s

Question 43Question →

Which property of transition metals primarily enables them to function as efficient catalysts in industrial chemical processes?

Ability to exhibit variable oxidation states and offer vacant d-orbitals

High electrical and thermal conductivity

Low melting points and low densities relative to s-block metals

Ability to shift the position of dynamic equilibrium to increase product yield

Show answer & explanation

Answer: Ability to exhibit variable oxidation states and offer vacant d-orbitals

Answer

Ability to exhibit variable oxidation states and offer vacant d-orbitals

Transition elements possess unfilled or partially filled d-orbitals and readily change oxidation states. These characteristics allow them to bind reactants temporarily or form unstable transition-state intermediates, offering reaction mechanisms with lower activation energy.

Step-by-Step Solution

1

Recall the fundamental feature of transition metals responsible for catalytic activity.

Transition metals have partially filled d-orbitals and can easily exchange electrons to adopt multiple oxidation states.

This allows them to form temporary bonds with reactants (adsorption) or form reactive intermediate species, which significantly lowers activation energy.

Key Concept

Catalytic behavior of transition metals due to variable oxidation states and available d-orbitals

Question 44Question →

Malachite is an important copper ore with the chemical formula

\text{CuCO}_3\cdot\text{Cu(OH)}_2

. Upon strong heating, it undergoes thermal decomposition according to the following balanced equation:

\text{CuCO}_3\cdot\text{Cu(OH)}_2(s) \rightarrow 2\text{CuO}(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)

What is the mass of copper(II) oxide (

\text{CuO}

), in grams, formed when

22.2\text{ g}

of malachite is completely decomposed? [Relative atomic masses:

\text{Cu} = 64

,

\text{C} = 12

,

\text{O} = 16

,

\text{H} = 1

]

Show answer & explanation

Answer: 16

Answer

The mass of copper(II) oxide produced is 16.0 g.

Thermal decomposition of 1 mole of malachite (

\text{CuCO}_3\cdot\text{Cu(OH)}_2

, molar mass

222\text{ g mol}^{-1}

) produces 2 moles of copper(II) oxide (

\text{CuO}

, molar mass

80\text{ g mol}^{-1}

). Given

22.2\text{ g}

of malachite (

0.1\text{ mol}

), exactly

0.2\text{ mol}

of

\text{CuO}

is formed, giving a mass of

0.2\text{ mol} \times 80\text{ g mol}^{-1} = 16.0\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of malachite,

\text{CuCO}_3\cdot\text{Cu(OH)}_2

222\text{ g mol}^{-1}

Summing the relative atomic masses:

2(64) + 12 + 5(16) + 2(1) = 222\text{ g mol}^{-1}

.

2

Calculate the number of moles of malachite in the sample

0.1\text{ mol}

Dividing given mass by molar mass:

\frac{22.2\text{ g}}{222\text{ g mol}^{-1}} = 0.1\text{ mol}

.

3

Determine the moles of

\text{CuO}

produced using stoichiometry

0.2\text{ mol}

of

\text{CuO}

The balanced chemical equation shows a

1:2

mole ratio between malachite and

\text{CuO}

.

4

Calculate the mass of

\text{CuO}

formed

16.0\text{ g}

Multiplying moles of

\text{CuO}

by its molar mass (

80\text{ g mol}^{-1}

):

0.2 \times 80 = 16.0\text{ g}

.

Key Concept

Stoichiometry of copper compounds thermal decomposition

Question 45Question →

Complete the statement regarding the transition metal complex ion $[Fe(CN)_6]^{3-}$ by filling in the missing values.

Fill in the blanks below

In the hexacyanoferrate(III) complex ion,

[Fe(CN)_6]^{3-}

, the coordination number of the central iron ion is and the oxidation state of iron is .

Show answer & explanation

Answer

The coordination number of the central iron ion in

[Fe(CN)_6]^{3-}

is 6, and its oxidation state is +3.

In

[Fe(CN)_6]^{3-}

, iron is bonded to six cyano (

CN^-

) ligands, establishing a coordination number of 6. Balancing the total charge gives

x + 6(-1) = -3

, yielding an oxidation state of +3 for the iron central metal ion.

Step-by-Step Solution

1

Determine the coordination number from the formula of the complex ion.

The central iron ion is bonded to 6 monodentate cyano (

CN^-

) ligands, so the coordination number is 6.

The coordination number represents the total number of ligand donor atoms attached directly to the central transition metal ion.

2

Calculate the oxidation state of the central iron ion.

Let

x

be the oxidation state of

Fe

. Each cyano ligand has a charge of

-1

, and the total complex ion charge is

-3

. Setting up the equation:

x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3

.

The sum of the oxidation state of the central metal and the charges of all ligands equals the net charge of the complex ion.

Key Concept

Coordination Number and Oxidation State Determination in Complex Ions

Question 46Question →

During the industrial extraction of iron in a blast furnace, limestone ( $\text{CaCO}_3$ ) is added to remove silica impurities ( $\text{SiO}_2$ ). Arrange the following stages of slag formation and separation in their correct chronological sequence from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence is: Thermal decomposition of limestone (

\text{CaCO}_3

) into basic calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

)

\rightarrow

Acid-base reaction between basic calcium oxide (

\text{CaO}

) flux and acidic silicon(IV) oxide (

\text{SiO}_2

) impurity

\rightarrow

Formation of molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

)

\rightarrow

Collection of molten slag as a protective layer floating above the denser liquid pig iron at the hearth.

Limestone (

\text{CaCO}_3

) decomposes thermally under high temperatures into calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

). Next, the basic

\text{CaO}

flux reacts with acidic silica (

\text{SiO}_2

) impurities in an acid-base neutralization to form molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

, slag). Finally, the molten slag collects at the hearth and floats on top of the denser molten pig iron, forming a protective layer.

Step-by-Step Solution

1

Identify the initial thermal reaction of limestone in the furnace.

Limestone (

\text{CaCO}_3

) decomposes at around

800^\circ\text{C}-1000^\circ\text{C}

to yield calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

).

Calcium oxide (

\text{CaO}

) must first be synthesized to act as a basic flux.

2

Determine the chemical interaction between the flux and raw ore impurities.

Basic calcium oxide (

\text{CaO}

) reacts with acidic silicon(IV) oxide (

\text{SiO}_2

).

Silica is the main acidic impurity in hematite ore, requiring neutralization by the basic flux.

3

Identify the chemical product formed from this reaction.

Molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

, slag) is formed via the reaction

\text{CaO(s)} + \text{SiO}_2\text{(s)} \rightarrow \text{CaSiO}_3\text{(l)}

.

Neutralization of silica forms the molten compound known as slag.

4

Describe the physical separation of slag at the hearth.

The molten slag drains down to the hearth and floats above the denser liquid iron layer.

Density differences allow slag to float on molten iron, preventing its re-oxidation by blast air.

Key Concept

Slag Formation and Impurity Removal in the Blast Furnace

Question 47Question →

What is the oxidation state of the central iron atom in the hexacyanoferrate(III) complex ion, $[Fe(CN)_6]^{3-}$ ?

+3

+2

+6

-3

Show answer & explanation

Answer: $+3$

Answer

The oxidation state of the central iron atom in

[Fe(CN)_6]^{3-}

is

+3

.

In the complex ion

[Fe(CN)_6]^{3-}

, six anionic cyanide ligands (

CN^-

), each with a

-1

charge, bond to the central iron atom. The sum of the oxidation state of iron (

x

) and the total ligand charge (

6 \times -1 = -6

) equals the overall charge of the complex ion (

-3

). Solving

x - 6 = -3

gives

x = +3

.

Step-by-Step Solution

1

Identify the charge of the ligands and the net charge of the complex ion.

Each cyanide ligand (

CN^-

) carries a

-1

charge. The overall complex ion has a net charge of

-3

.

Ligand charges must sum with the central metal oxidation state to equal the overall ion charge.

2

Set up an algebraic equation for the oxidation state of iron (

x

).

x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3

.

Solving the linear algebraic equation yields the precise oxidation state of the central atom.

Key Concept

Determination of central metal oxidation state in complex ions

Estimated Time:45s

Question 48Question →

Which of the following transition metal compounds is used as the catalyst in the industrial manufacture of tetraoxosulfate(VI) acid by the Contact process?

Vanadium(V) oxide (

V_2O_5

)

Finely divided iron (

Fe

)

Finely divided nickel (

Ni

)

Manganese(IV) oxide (

MnO_2

)

Show answer & explanation

Answer: Vanadium(V) oxide ( $V_2O_5$ )

Answer

Vanadium(V) oxide (

V_2O_5

) is the catalyst used in the Contact process.

Vanadium(V) oxide (

V_2O_5

) is the primary catalyst used in the Contact process to catalyze the reversible oxidation of

SO_2

to

SO_3

due to the ability of vanadium to vary its oxidation state between +5 and +4 during the catalytic cycle.

Step-by-Step Solution

1

Identify the key chemical transformation in the Contact process

The oxidation of sulfur(IV) oxide gas to sulfur(VI) oxide gas (

2SO_2 + O_2 \rightleftharpoons 2SO_3

).

This exothermic reversible step requires a catalyst to achieve an optimal reaction rate at reasonable temperature.

2

Recall the industrial catalyst specifically employed for this conversion

Vanadium(V) oxide (

V_2O_5

) is used as the preferred catalyst at around 450 °C.

Transition metals and their compounds exhibit variable oxidation states, allowing

V_2O_5

to facilitate electron transfer steps efficiently.

Key Concept

Catalytic behavior of transition metal compounds in industrial processes

Estimated Time:45s

Question 49Question →

When iron filings are heated in a stream of dry chlorine gas, compound $X$ is formed. Conversely, when iron filings react with dilute hydrochloric acid, compound $Y$ is produced. What are the correct IUPAC names of compounds $X$ and $Y$ respectively?

Iron(III) chloride and iron(II) chloride

Iron(II) chloride and iron(III) chloride

Iron(III) chloride and iron(III) chloride

Iron(II) chloride and iron(II) chloride

Show answer & explanation

Answer: Iron(III) chloride and iron(II) chloride

Answer

Iron(III) chloride and iron(II) chloride

Dry chlorine gas acts as a powerful oxidizing agent that oxidizes iron directly to iron(III) chloride (

\text{FeCl}_3

). In contrast, reacting iron with dilute hydrochloric acid generates iron(II) chloride (

\text{FeCl}_2

) because the evolved hydrogen gas acts as a reducing agent, maintaining iron in the

+2

oxidation state.

Step-by-Step Solution

1

Analyze the reaction of iron with dry chlorine gas

2\text{Fe}_{(s)} + 3\text{Cl}_{2(g)} \rightarrow 2\text{FeCl}_{3(s)}

Chlorine gas is a powerful oxidizing agent capable of taking iron from oxidation state 0 to +3, forming iron(III) chloride.

2

Analyze the reaction of iron with dilute hydrochloric acid

\text{Fe}_{(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{FeCl}_{2(aq)} + \text{H}_{2(g)}

Dilute hydrochloric acid oxidizes iron to iron(II) chloride (

+2

state). The hydrogen gas (

\text{H}_2

) evolved during the reaction acts as a reducing agent, preventing any further oxidation to iron(III).

3

Match the products to compound

X

and compound

Y

Compound

X

is iron(III) chloride and compound

Y

is iron(II) chloride.

Sequential identification following the given stem conditions.

Key Concept

Variable oxidation states of iron and differential oxidizing strength of chlorine versus hydrogen ions.

Estimated Time:1m 0s

Question 50Question →

A metallurgical research laboratory analyzed three different metallic alloys ( $P$ , $Q$ , and $R$ ) to evaluate their internal lattice structures and physical property modifications relative to their primary base metals:

- Sample $P$ : Consists of copper with $30\%$ zinc solute atoms.
- Sample $Q$ : Consists of iron with $1.0\%$ carbon solute atoms.
- Sample $R$ : Consists of aluminium alloyed with small amounts of copper, magnesium, and manganese.

Which of the following statements correctly classifies the lattice alloy types of $P$ , $Q$ , and $R$ and accurately describes their electrical conductivity relative to their pure base metals?

Sample

P

and Sample

R

are substitutional alloys, while Sample

Q

is an interstitial alloy; all three alloys exhibit lower electrical conductivity than their pure base metals due to lattice distortion.

Sample

P

and Sample

Q

are interstitial alloys, while Sample

R

is a substitutional alloy; all three alloys exhibit higher electrical conductivity than their pure base metals due to enhanced metallic bonding.

Sample

Q

is a substitutional alloy, while Sample

P

and Sample

R

are interstitial alloys; Sample

Q

exhibits lower electrical conductivity, whereas Sample

P

and Sample

R

exhibit unchanged conductivity.

Sample

P

, Sample

Q

, and Sample

R

are all interstitial alloys; their electrical conductivity remains higher than that of pure copper, iron, and aluminium respectively because solute atoms donate free electrons to the delocalized sea.

Show answer & explanation

Answer: Sample $P$ and Sample $R$ are substitutional alloys, while Sample $Q$ is an interstitial alloy; all three alloys exhibit lower electrical conductivity than their pure base metals due to lattice distortion.

Answer

Sample P and Sample R are substitutional alloys, while Sample Q is an interstitial alloy; all three alloys exhibit lower electrical conductivity than their pure base metals due to lattice distortion.

The correct option accurately identifies that brass (Sample P) and duralumin (Sample R) are substitutional alloys because their solute atoms have comparable atomic radii to their host atoms, whereas carbon steel (Sample Q) is an interstitial alloy because carbon atoms are small enough to enter the voids of the iron lattice. Furthermore, it correctly states that all alloys suffer a reduction in electrical conductivity relative to pure host metals due to electron scattering caused by lattice distortion.

Step-by-Step Solution

1

Determine the alloy lattice type for Sample P (brass) and Sample R (duralumin).

Zinc atoms (in brass) and copper/magnesium/manganese atoms (in duralumin) have atomic radii within

15\%

of the host copper and aluminium radii, so they replace host atoms directly in the crystal lattice, forming substitutional alloys.

When solute and solvent atomic radii are comparable, solute atoms substitute for host metal atoms at lattice sites.

2

Determine the alloy lattice type for Sample Q (carbon steel).

Carbon has a significantly smaller atomic radius than iron, allowing carbon atoms to fit into the interstitial voids between iron atoms, forming an interstitial alloy.

Solute atoms with radii much smaller than the host metal fit into spaces (interstices) between lattice atoms.

3

Evaluate the effect of alloying on electrical conductivity.

The presence of foreign solute atoms (whether substitutional or interstitial) causes lattice distortion, which disrupts the uniform periodic potential of the metal lattice and scatters conduction electrons, resulting in lower electrical conductivity compared to pure base metals.

Electron mobility is reduced by lattice irregularities and strain fields created by solute atoms.

Key Concept

Classification of substitutional vs interstitial alloys and the effect of lattice distortion on physical properties like electrical conductivity.

Estimated Time:2m 0s

Question 51Question →

A sample of hydrated copper(II) tetraoxosulfate(VI), $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$ , with a mass of $12.5\text{ g}$ is dissolved completely in distilled water to make $250\text{ cm}^3$ of solution. What is the concentration of the copper(II) tetraoxosulfate(VI) solution in $\text{mol dm}^{-3}$ ?

( $\text{Relative atomic masses: Cu} = 64,\text{ S} = 32,\text{ O} = 16,\text{ H} = 1$ )

0.05\text{ mol dm}^{-3}

0.20\text{ mol dm}^{-3}

0.31\text{ mol dm}^{-3}

50.00\text{ mol dm}^{-3}

Show answer & explanation

Answer: $0.20\text{ mol dm}^{-3}$

Answer

The molar concentration of the copper(II) tetraoxosulfate(VI) solution is

0.20\text{ mol dm}^{-3}

.

The correct concentration is

0.20\text{ mol dm}^{-3}

. The molar mass of hydrated copper(II) tetraoxosulfate(VI) (

\text{CuSO}_4\cdot 5\text{H}_2\text{O}

) is

250\text{ g mol}^{-1}

, so

12.5\text{ g}

corresponds to

0.05\text{ mol}

. Dividing

0.05\text{ mol}

by the volume of

0.25\text{ dm}^3

gives

0.20\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Calculate the molar mass of hydrated copper(II) tetraoxosulfate(VI),

\text{CuSO}_4\cdot 5\text{H}_2\text{O}

.

\text{Molar mass} = 64 + 32 + (4 \times 16) + 5 \times (2 \times 1 + 16) = 250\text{ g mol}^{-1}

.

The total molar mass must include the 5 molecules of water of crystallization present in the solid salt.

2

Determine the number of moles of

\text{CuSO}_4\cdot 5\text{H}_2\text{O}

dissolved.

\text{Moles} = \frac{12.5\text{ g}}{250\text{ g mol}^{-1}} = 0.05\text{ mol}

.

Number of moles is equal to the given mass divided by the molar mass.

3

Convert the volume of the solution from

\text{cm}^3

to

\text{dm}^3

.

\text{Volume} = \frac{250\text{ cm}^3}{1000} = 0.25\text{ dm}^3

.

Molar concentration requires the volume of solution to be expressed in cubic decimetres.

4

Calculate the molar concentration in

\text{mol dm}^{-3}

.

\text{Concentration} = \frac{0.05\text{ mol}}{0.25\text{ dm}^3} = 0.20\text{ mol dm}^{-3}

.

Molarity is defined as moles of solute per cubic decimetre of solution.

Key Concept

Molar Concentration of Hydrated Copper Compounds

Estimated Time:1m 30s

Question 52Question →

Chromium is a first-row transition element with an atomic number of 24. During a chemical reaction, a neutral chromium atom loses three electrons to form the chromium(III) cation, $Cr^{3+}$ . Which electronic structure correctly represents this $Cr^{3+}$ ion in its ground state?

[Ar] 3d^3

[Ar] 3d^1 4s^2

[Ar] 3d^2 4s^1

[Ar] 3d^4

Show answer & explanation

Answer: $[Ar] 3d^3$

Answer

The ground-state electronic configuration of the chromium(III) ion,

Cr^{3+}

, is

[Ar] 3d^3

.

Neutral chromium has an electronic configuration of

[Ar] 3d^5 4s^1

due to half-filled subshell stability. When forming a

Cr^{3+}

cation, the atom loses three electrons: the single valence electron in the

4s

orbital is lost first, followed by two electrons from the

3d

subshell, leaving a final ground-state configuration of

[Ar] 3d^3

.

Step-by-Step Solution

1

Determine the ground-state electronic configuration of neutral chromium (

Cr

,

Z = 24

).

Neutral chromium has the anomalous configuration

[Ar] 3d^5 4s^1

.

Chromium exhibits an exception to the standard Aufbau principle because a half-filled

d

-subshell (

3d^5

) provides extra thermodynamic stability.

2

Apply the rule for cation formation in transition metals.

Electrons in the outermost principal quantum shell (

4s

) are removed prior to removing electrons from the inner

(n-1)d

subshell (

3d

).

Once filled, the

3d

orbitals experience greater nuclear attraction and drop lower in energy than the

4s

orbital.

3

Deduct three electrons to account for the

+3

charge of

Cr^{3+}

.

Remove the single electron from

4s

(

[Ar] 3d^5 4s^0

), then remove two electrons from

3d

to obtain

[Ar] 3d^3

.

Removing 3 electrons total converts neutral

Cr

into

Cr^{3+}

.

Key Concept

Electronic Configuration of Transition Metal Cations

Estimated Time:2m 0s

Question 53Question →

Match each substance or chemical reagent related to iron extraction, corrosion, and qualitative analysis on the left with its corresponding chemical role or property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Coke matches with reacting with CO2 to generate CO; Calcium silicate matches with forming molten slag to prevent re-oxidation; Hydrated iron(III) oxide matches with the chemical composition of rust; Potassium hexacyanoferrate(III) matches with testing for Fe(II) ions.

Each substance is matched directly to its chemical function: Coke supplies carbon to generate carbon(II) oxide gas; Calcium silicate acts as slag floating atop molten iron; Hydrated iron(III) oxide is the exact chemical composition of rust; Potassium hexacyanoferrate(III) is the standard bench reagent for detecting iron(II) ions in qualitative testing.

Step-by-Step Solution

1

Analyze the blast furnace chemical reactions involving carbon input.

Coke (

\text{C}

) reacts with ascending

\text{CO}_2

(

\text{C} + \text{CO}_2 \rightarrow 2\text{CO}

) to produce carbon(II) oxide, which acts as the chief reducing agent for haematite.

Identify the role of Coke in blast furnace extraction.

2

Analyze slag formation and its function in the blast furnace.

Lime (

\text{CaO}

) combines with silica (

\text{SiO}_2

) to yield calcium silicate (

\text{CaSiO}_3

), a molten waste slag that floats above molten iron.

Identify the function of calcium silicate in molten iron isolation.

3

Identify the chemical identity of rust.

Atmospheric corrosion of iron in the presence of oxygen and water forms reddish-brown hydrated iron(III) oxide (

\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}

).

Match the rust formula with its physical phenomenon.

4

Recall qualitative analysis reagents for iron oxidation states.

Potassium hexacyanoferrate(III) reacts with

\text{Fe}^{2+}

ions to give a characteristic dark blue precipitate.

Match the analytical reagent with its specific ion test.

Key Concept

Extraction of Iron in the Blast Furnace, Rusting Mechanism, and Qualitative Analysis of Iron Ions

Question 54Question →

Brass is an alloy widely utilized in making musical instruments, door handles, and decorative fittings due to its strength, acoustic properties, and resistance to corrosion. Which pair of metals constitutes brass?

Copper and zinc

Copper and tin

Lead and tin

Iron and carbon

Show answer & explanation

Answer: Copper and zinc

Answer

Copper and zinc

Brass is an alloy specifically produced by melting together copper and zinc. The addition of zinc improves the mechanical strength and workability relative to pure copper.

Step-by-Step Solution

1

Identify the standard elemental composition of the specified metallic alloy.

Brass is a substitutional metallic alloy composed primarily of copper (Cu) and zinc (Zn).

Alloys are mixtures of metals or a metal with another element; brass specifically pairs copper as the primary metal with zinc.

Key Concept

Elemental compositions of common copper alloys (brass vs bronze)

Estimated Time:45s

Question 55Question →

During the industrial extraction of iron in a blast furnace, hematite ( $\text{Fe}_2\text{O}_3$ ) is reduced to molten iron in the upper reduction zone. Which chemical species serves as the chief reducing agent responsible for this reduction?

Carbon(II) oxide (

\text{CO}

)

Solid carbon (coke,

\text{C}

)

Calcium oxide (

\text{CaO}

)

Carbon(IV) oxide (

\text{CO}_2

)

Show answer & explanation

Answer: Carbon(II) oxide ( $\text{CO}$ )

Answer

Carbon(II) oxide (

\text{CO}

) is the chief reducing agent in the upper reduction zone of the blast furnace.

In the upper, lower-temperature region of the blast furnace, ascending carbon(II) oxide (

\text{CO}

) gas chemically reduces hematite (

\text{Fe}_2\text{O}_3

) to iron, producing carbon(IV) oxide (

\text{CO}_2

) gas as a byproduct.

Step-by-Step Solution

1

Analyze the reactions occurring in the upper temperature zone (

400^\circ\text{C} - 700^\circ\text{C}

) of the blast furnace.

Ascending carbon(II) oxide gas contacts descending solid hematite ore.

Gaseous carbon(II) oxide provides effective gas-solid contact necessary for chemical reduction.

2

Write the balanced chemical equation for the reduction of hematite.

\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(l) + 3\text{CO}_2(g)

Carbon(II) oxide removes oxygen from iron(III) oxide, reducing iron from the +3 oxidation state to metallic iron (0 oxidation state).

Key Concept

Iron Extraction in the Blast Furnace

Question 56Question →

In the auto-reduction stage of copper extraction, copper(I) oxide (

\text{Cu}_2\text{O}

) reacts with copper(I) sulfide (

\text{Cu}_2\text{S}

) according to the following balanced equation:

\text{Cu}_2\text{S (s)} + 2\text{Cu}_2\text{O (s)} \rightarrow 6\text{Cu (s)} + \text{SO}_2\text{ (g)}

If

14.3\text{ g}

of copper(I) oxide reacts completely with excess copper(I) sulfide, what mass of metallic copper in grams is produced?

[\text{Cu} = 63.5, \text{O} = 16.0]

Show answer & explanation

Answer: 19.05

Answer

The mass of metallic copper produced is 19.05 g.

According to the balanced chemical equation, 2 moles of copper(I) oxide react with copper(I) sulfide to yield 6 moles of metallic copper, which simplifies to a 1:3 molar ratio. Given that the molar mass of Cu2O is 143 g/mol, 14.3 g represents 0.1 mol of Cu2O. Based on the 1:3 ratio, this produces 0.3 mol of copper metal, corresponding to 19.05 g of Cu.

Step-by-Step Solution

1

Calculate the molar mass of copper(I) oxide

Molar mass of Cu2O = 143.0 g/mol

Required to convert the given mass of reactant to moles.

2

Determine moles of Cu2O reacted

Moles of Cu2O = 14.3 g / 143.0 g/mol = 0.10 mol

Establishes the quantitative amount of Cu2O in the chemical system.

3

Determine moles of copper metal produced using stoichiometry

Moles of Cu = 0.10 mol * (6 / 2) = 0.30 mol

The balanced chemical equation shows 2 moles of Cu2O yield 6 moles of Cu metal.

4

Calculate the mass of copper metal produced

Mass of Cu = 0.30 mol * 63.5 g/mol = 19.05 g

Converts the stoichiometric amount of product moles into grams.

Key Concept

Auto-reduction in copper extraction and stoichiometric mass calculations

Question 57Question →

Match each transition metal complex ion on the left with its corresponding structural, electronic, and magnetic characteristics on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

[Fe(CN)_6]^{3-}

matches Octahedral geometry with a

3d^5

low-spin central metal ion containing 1 unpaired electron;

[Ni(CN)_4]^{2-}

matches Square planar geometry with a

3d^8

central metal ion that is diamagnetic;

[Co(NH_3)_6]^{3+}

matches Octahedral geometry with a

3d^6

low-spin central metal ion that is diamagnetic;

[Cu(NH_3)_4]^{2+}

matches Square planar geometry with a

3d^9

central metal ion containing 1 unpaired electron.

Each complex ion's central metal ion exhibits a specific oxidation state, electronic configuration, coordination geometry, and spin state based on crystal field theory and ligand field strength.

[Fe(CN)_6]^{3-}

features

Fe^{3+}

(

3d^5

) in a low-spin octahedral state with 1 unpaired electron.

[Ni(CN)_4]^{2-}

features

Ni^{2+}

(

3d^8

) in a square planar diamagnetic configuration.

[Co(NH_3)_6]^{3+}

features

Co^{3+}

(

3d^6

) in a low-spin octahedral diamagnetic state.

[Cu(NH_3)_4]^{2+}

features

Cu^{2+}

(

3d^9

) in a square planar configuration with 1 unpaired electron.

Step-by-Step Solution

1

Determine the oxidation state and d-electron count of the central metal ion in each complex ion.

For

[Fe(CN)_6]^{3-}

,

Fe^{3+}

is

3d^5

. For

[Ni(CN)_4]^{2-}

,

Ni^{2+}

is

3d^8

. For

[Co(NH_3)_6]^{3+}

,

Co^{3+}

is

3d^6

. For

[Cu(NH_3)_4]^{2+}

,

Cu^{2+}

is

3d^9

.

Ligand charges (

CN^-

=

-1

,

NH_3

=

0

) determine the oxidation state of the central transition metal ion.

2

Analyze ligand strength, coordination geometry, and crystal field splitting to determine magnetic character.

[Fe(CN)_6]^{3-}

is octahedral low-spin (

t_{2g}^5

, 1 unpaired

e^-

).

[Ni(CN)_4]^{2-}

is square planar (

dsp^2

, diamagnetic).

[Co(NH_3)_6]^{3+}

is octahedral low-spin (

t_{2g}^6

, diamagnetic).

[Cu(NH_3)_4]^{2+}

is square planar (

3d^9

, 1 unpaired

e^-

).

Strong-field ligands (

CN^-

,

NH_3

) induce electron pairing in low-spin octahedral or square planar configurations.

3

Match each complex ion to its complete set of physical, electronic, and magnetic properties.

Each complex correctly aligns with its unique d-electron configuration, geometry, and spin state.

Verifies all coordination parameters systematically.

Key Concept

Electronic Configuration, Oxidation State, Geometry, and Magnetic Properties of Transition Metal Complexes

Question 58Question →

Match each aluminium alloy or chemical substance in Column I with its primary industrial composition or application in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Duralumin matches with High-strength alloy of aluminium, copper, magnesium, and manganese used in aircraft structural frames; Magnalium matches with Lightweight corrosion-resistant alloy of aluminium and magnesium used in balance beams and aircraft parts; Alnico matches with Alloy of aluminium, nickel, cobalt, and iron utilized in manufacturing strong permanent magnets; Molten Cryolite matches with Molten solvent added to lower the operating temperature of alumina and improve electrical conductivity.

Each substance is correctly paired based on standard chemistry principles: Duralumin is an

Al-Cu-Mg-Mn

structural aircraft alloy, Magnalium is an

Al-Mg

lightweight alloy, Alnico is an

Al-Ni-Co-Fe

magnetic alloy, and molten cryolite (

Na_3AlF_6

) lowers the melting temperature of alumina during electrolysis.

Step-by-Step Solution

1

Identify the chemical composition and primary application of Duralumin.

Duralumin contains

Al

,

Cu

,

Mg

, and

Mn

, known for its structural strength in aircraft manufacture.

Copper adds tensile strength to aluminium while retaining low density.

2

Identify the composition and application of Magnalium.

Magnalium is an alloy of

Al

and

Mg

, valued for low density and high corrosion resistance.

Magnesium enhances hardness and lightness without increasing susceptibility to oxidation.

3

Determine the composition and use of Alnico.

Alnico consists of

Al

,

Ni

,

Co

, and

Fe

, used for permanent magnets.

Ferromagnetic elements combined with aluminium create high magnetic retentivity.

4

Determine the role of cryolite in the industrial extraction of aluminium.

Cryolite acts as an electrolytic solvent, lowering the melting point of

Al_2O_3

and increasing conductivity.

Pure alumina has an extremely high melting point (

2050^\circ\text{C}

); dissolving it in molten cryolite reduces energy consumption.

Key Concept

Industrial extraction of aluminium and compositions/applications of its major alloys

Question 59Question →

Alloys are frequently used in structural engineering and manufacturing because they exhibit greater mechanical strength and hardness than pure metals. Which of the following structural factors best explains why introducing a secondary element increases the hardness of a metal?

Atoms of differing sizes disrupt the regular crystal lattice, hindering the movement of atomic layers past one another.

The constituent elements undergo a chemical reaction to form rigid giant covalent networks throughout the solid.

Delocalized valence electrons are completely removed, converting the metallic lattice into an ionic crystal structure.

The added atoms expand in volume to completely fill all interatomic spaces and stop thermal vibrations.

Show answer & explanation

Answer: Atoms of differing sizes disrupt the regular crystal lattice, hindering the movement of atomic layers past one another.

Answer

Atoms of differing sizes disrupt the regular crystal lattice, hindering the movement of atomic layers past one another.

In a pure metal, identical atoms form regular layers that slip over one another with relative ease. When a secondary element with a different atomic radius is introduced to form an alloy, the regular lattice arrangement is distorted. This disruption prevents atomic layers from sliding smoothly, significantly increasing the hardness and tensile strength of the alloy.

Step-by-Step Solution

1

Consider the structure of a pure metal.

In pure metals, identical atoms are arranged in uniform, orderly layers that easily slide over each other under shear stress, making the metal malleable and relatively soft.

Uniform atomic radii allow smooth plane slipping along lattice planes.

2

Analyze the structural effect of alloying.

Adding atoms of a different size introduces structural irregularities and distorts the host crystal lattice.

The distortion creates friction and resistance against the sliding of atomic layers, resulting in enhanced hardness and strength.

Key Concept

Lattice distortion and mechanical strengthening of alloys

Estimated Time:45s

Question 60Question →

Match each specialized metallic alloy listed on the left with its correct elemental composition and primary functional application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair German Silver with the copper-zinc-nickel silver-free alloy, Alnico with the aluminium-nickel-cobalt-iron magnetic alloy, Type Metal with the lead-tin-antimony alloy that expands upon solidification, and Wood's Metal with the low-melting bismuth-lead-tin-cadmium fusible alloy.

Each alloy aligns strictly with its chemical formula and specialized physical property: German Silver (

\text{Cu}-\text{Zn}-\text{Ni}

) is silver-free with high resistance; Alnico (

\text{Al}-\text{Ni}-\text{Co}-\text{Fe}

) forms permanent magnets; Type Metal (

\text{Pb}-\text{Sn}-\text{Sb}

) expands upon freezing; Wood's Metal (

\text{Bi}-\text{Pb}-\text{Sn}-\text{Cd}

) melts at

65\ ^\circ\text{C}

.

Step-by-Step Solution

1

Analyze the chemical composition and key characteristic of German Silver.

German Silver consists of

\text{Cu}

,

\text{Zn}

, and

\text{Ni}

without any silver content, valued for high electrical resistivity and silvery appearance.

Identifying naming misnomers prevents confusing German Silver with silver-bearing precious metal alloys.

2

Deconstruct the constituent components and physical property of Alnico.

Alnico combines

\text{Al}

,

\text{Ni}

,

\text{Co}

, and

\text{Fe}

to form hard ferromagnetic structures.

The alloy acronym highlights its elements (

\text{Al}-\text{Ni}-\text{Co}

), which deliver superior permanent magnetic strength.

3

Examine the solid-phase volume change characteristic of Type Metal.

Type Metal contains

\text{Pb}

,

\text{Sn}

, and

\text{Sb}

; the presence of antimony induces volumetric expansion upon cooling.

While most metals shrink when freezing, antimony forces Type Metal to expand into fine matrix details during printing press type manufacture.

4

Evaluate the thermal melting point and application of Wood's Metal.

Wood's Metal is a eutectic combination of

\text{Bi}

,

\text{Pb}

,

\text{Sn}

, and

\text{Cd}

melting at

65\ ^\circ\text{C}

.

Combining four metals in specific proportions disrupts individual crystal lattice stability, dropping the melting point below

100\ ^\circ\text{C}

for safety sprinkler valves.

Key Concept

Alloy Classifications, Compositions, Phase-Change Properties, and Industrial Uses

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